Elijah, Eric and Ethan shared a bag of coins. Eric took 35 as many coins as Ethan. Elijah took twice as many coins as the total Eric and Ethan took. After Elijah had given 29 to Eric and 3 to Ethan, Eric gave 9 to Ethan. In the end, all three of them had the same number of coins. Find the difference between the number of coins that Elijah and Eric had at first.
| |
Eric |
Ethan |
Elijah |
Total |
| Before |
3 u |
5 u |
16 u |
24 u |
| Change 1 |
+ 29 |
|
- 29 |
|
| Change 2 |
|
+ 3 |
- 3 |
|
| Change 3 |
- 9 |
+ 9 |
|
|
| After |
1x8 =
8 u |
1x8 =
8 u |
1x8 =
8 u |
3x8 =
24 u |
Total number of coins that Eric and Ethan had at first
= 3 u + 5 u
= 8 u
Number of coins that Elijah had at first
= 2 x 8 u
= 16 u
The total number of coins remains unchanged. Make the total number of coins the same. LCM of 3 and 24 is 24.
Number of coins that Ethan received from Eric and Elijah
= 8 u - 5 u
= 3 u
3 u = 3 + 9
3 u = 12
1 u = 12 ÷ 3 = 4
Difference between the number coins that Elijah and Eric had at first
= 16 u - 3 u
= 13 u
= 13 x 4
= 52
Answer(s): 52
